Fixed points of finite order mapping classes I am currently working myself through the proof that every finite order element of the mapping class group $\mathrm{MCG}(S_g)=\mathrm{Hom}^+(S_g) / \mathrm{Hom}_0(S_g)$ of a closed, hyperbolic, genus $g$ surface has a fixed point in Teichmüller space. The reference I am using is Farb and Margalit's "A Primer on Mapping Class Groups", page 371. They use the following argument that I do not understand:
Suppose $f \in \mathrm{MCG}(S_g)$ has order $n < \infty$ and consider the action of the finite cycic subgroup $\langle f \rangle < \mathrm{MCG}(S_g)$ on Teichmüller space. They write: "The finite cyclic group $\langle f \rangle$ cannot act freely on Teichmüller space for otherwise the quotient [Teich/$\langle f \rangle$] would be a finite-dimensional $K(\mathbb{Z/nZ},1)$ [i.e. a space with fundamental group isomorphic to $\mathbb{Z/nZ}$]. Thus, $f^k \cdot X = X$ for some $1 \leq k < n$ and some $X$ in Teichmüller space."
I understand why the action being free implies that the quotient is such a space but I do not understand why that is a contradiction.
Cheers!
 A: The Teichmüller space of a genus $g$ surface is homeomorphic to a ball of dimension $6g - 6$, in particular, it is contractible. If $\mathbb{Z}/n\mathbb{Z}$ acted freely on it, the quotient would be a space (in fact a manifold of dimension $6g-6$) with fundamental group $\mathbb{Z}/n\mathbb{Z}$ and contractible universal cover, so all its higher homotopy groups vanish. Therefore, the quotient is an Eilenberg-MacLane space, namely a $K(\mathbb{Z}/n\mathbb{Z}, 1)$. 
Another example of a $K(\mathbb{Z}/n\mathbb{Z}, 1)$ is the infinite Lens space $S^{\infty}/(\mathbb{Z}/n\mathbb{Z})$ which has non-zero cohomology in infinitely many degrees. As $K(G, m)$'s are unique up to homotopy, the same is true of the quotient of Teichmüller space. But the quotient is a $(6g - 6)$-dimensional manifold, so its cohomology vanishes in degrees larger than $6g - 6$. Therefore the quotient cannot be homotopy equivalent to $S^{\infty}/(\mathbb{Z}/n\mathbb{Z})$, so $\mathbb{Z}/n\mathbb{Z}$ does act freely on Teichmüller space.
